##### Homogenous Second Order Differential Equations

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I'm having a bit of difficulty obtaining the answer given at the back of my paper.

Here is the problem:

(d^2y/dt^2) - ( 6dy / dt ) + 8y = 0 our initial conditions: y(0) = 1 y'(0) = 0

working: -> m^2 - 6m + 8 = 0 put into formula:

m = (-b +- squareroot( b^2 - (4)ac)) / 2a working: -> m = 6 +- squareroot( -6^2 - (4)(1)^2(8)) / (2)(1) -> m = (6 +- squareroot( 36 - 32)) / 2

Can now identify the nature of our roots, which is Distinct so we're using this formula (k1 is a constant):

y(t) = K1e^(m1t) + k2e^(m2t) now continue our working from above to find m1 and m2:

-> (m = 6 + squareroot(4)) / 2 -> m1 = 4 -> m2 = 2 put m1 and m2 into the formula: y(t) = k1e^(4t) + k2e^(2t) Now this is the part that I don't understand. We now have to find the constants and all the answer in the book gives is:

y(0) k1 + k2 = 1 y'(0) = 2k1 + 4k2 = 0 solving: y(t) = 2e^(2t) - e^(4t)

Can someone explain how I get to this part with working please?

Aug 7th, 2015

Hello!

As I understand, you want to determine k1 and k2. We find them from the conditions y(0)=1 and y'(0)=0.

y(t)=k1*e^(4t) + k2*e^(2t),

y(0)=k1+k2=1,

y'(t)=4*k1*e^(4t) + 2*k2*e^(2t),

y'(0) = 4k1 + 2k2 = 0.

This is a linear system for k1 and k2, and it is simple to solve it.

k2 = 1 - k1, 4k1 + 2*(1-k1) = 0,

2k1 = -2, k1 = -1.

And k2=1-k1=2.

Aug 7th, 2015

Thank you very much for your help!  I understand now how you managed to get the constants. My book is quite unclear how it obtains these answers and I've been kind of thrown into the deep end with no Calculus experience.

Just leaves the final piece, which is where you take the constants K1, and K2 and put them into the original equation which gives:

y(t) = 2*e^(2*t) - e^(4*t)

Aug 7th, 2015

You are welcome:)

Aug 7th, 2015

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Aug 7th, 2015
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Aug 7th, 2015
Dec 6th, 2016
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